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HariniParthiban
Regular Visitor

RLS to filter data in a table from a set of data on another table

Hi,

I have to filter the report based on the branches the user has access to. I have a table called BRANCHESLIST which carries the information of branch numbers and user emails.

HariniParthiban_0-1712926674741.png

ASSETSETUPVIEW is the table I want to filter. Using filter directly on RLS throws an error:

HariniParthiban_1-1712926805809.png

I created a measure 

FilteredTable =
SUMMARIZE(
    FILTER(
        BRANCHESLIST,
        BRANCHESLIST[EMAIL_FW] = USERPRINCIPALNAME()
    ),
    BRANCHESLIST[COSTCENTRE_FW]
)

and then used RLS which gives the following error:

HariniParthiban_2-1712926921376.png

How can I filter the table in this scenario?

1 ACCEPTED SOLUTION
Anonymous
Not applicable

Hi  @HariniParthiban ,

 

You can place the following code:

var UPN=USERPRINCIPALNAME()
var UserRow=
    CALCULATETABLE(
        'BRANCHESLIST',
        'BRANCHESLIST'[EMAAIL_FW]=UPN)
var Flag=
SELECTCOLUMNS(
    UserRow,'BRANCHESLIST'[COSTCENTER])
var Result=
FILTER(
'ASSETSETUPVIEW','ASSETSETUPVIEW'[COSTTCENTRE_FW] in Flag)
RETURN
IF(
    'ASSETSETUPVIEW'[COSTTCENTRE_FW] in Flag,TRUE(),FALSE())

vyangliumsft_0-1713161297780.png

vyangliumsft_1-1713161297781.png

 

Best Regards,

Liu Yang

If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

 

View solution in original post

1 REPLY 1
Anonymous
Not applicable

Hi  @HariniParthiban ,

 

You can place the following code:

var UPN=USERPRINCIPALNAME()
var UserRow=
    CALCULATETABLE(
        'BRANCHESLIST',
        'BRANCHESLIST'[EMAAIL_FW]=UPN)
var Flag=
SELECTCOLUMNS(
    UserRow,'BRANCHESLIST'[COSTCENTER])
var Result=
FILTER(
'ASSETSETUPVIEW','ASSETSETUPVIEW'[COSTTCENTRE_FW] in Flag)
RETURN
IF(
    'ASSETSETUPVIEW'[COSTTCENTRE_FW] in Flag,TRUE(),FALSE())

vyangliumsft_0-1713161297780.png

vyangliumsft_1-1713161297781.png

 

Best Regards,

Liu Yang

If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

 

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